The function y= (3x^3-2x+1)/(2x^2) has me confused. I am trying to figure out why this function has a horizontal asymptote when the degree of the numerator is larger than the denominator, which I thought was supposed to mean it has no horizontal asymptote?

 

 

Thanks

It doesn't have a horizontal asymptote. Try graphing it.

Well, then my wonderful textbook(sarcasm) has an error in it....It says the answer is y=3/2*x.

That's roughly the slope of the line you get when you graph the original function.

 

Check the directions again to see what it's asking for.

It says determine any horizontal, vertical or slant asymptotes. I think it may be the slant asymptote. My bad....

 

I figure I should ask my other question on this post as well.

 

This problem is from a Cliff's Book

The function (2x-5)/(x^2-4x-5) has a horizontal asymptote of y=2. This function seems like it violates the rules........When I worked it out I got that y=0 because the the numerator has a smaller power than the denominator?

You're right again. I graphed it and it approaches 0.

Well then, I guess I did find an error this time. Thanks for the help.